Answer:
a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m
Explanation:
This can be answered with either force analysis and kinematics, or work and energy.
a) Using force analysis, we can draw a free body diagram for the snowboarder. There are two forces: normal force perpendicular to the slope and gravity down.
Sum of the forces parallel to the slope:
∑F = ma
mg sin θ = ma
a = g sin θ
Therefore, the velocity at the bottom is:
v² = v₀² + 2a(x - x₀)
v² = (0)² + 2(9.8 sin 60°) (45.7 / sin 60° - 0)
v = 29.9 m/s
Alternatively, using energy:
PE = KE
mgh = 1/2 mv²
v = √(2gh)
v = √(2×9.8×45.7)
v = 29.9 m/s
b) Drawing a free body diagram, there are three forces on the snowboarder. Normal force up, gravity down, and friction to the left.
Sum of the forces in the y direction:
∑F = ma
N - mg = 0
N = mg
Sum of the forces in the x direction:
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Therefore, the snowboarder's final speed is:
v² = v₀² + 2a(x - x₀)
v² = (29.9)² + 2(-9.8×.102) (10 - 0)
v = 29.6 m/s
Using energy instead:
KE = KE + W
1/2 mv² = 1/2 mv² + F d
1/2 mv² = 1/2 mv² + mgμ d
1/2 v² = 1/2 v² + gμ d
1/2 (29.9)² = 1/2 v² + (9.8)(0.102)(10)
v = 29.6 m/s
c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.
∑F = ma
-mg sin θ = ma
a = -g sin θ
Therefore, the final height reached is:
v² = v₀² + 2a(x - x₀)
(0)² = (29.6)² + 2(-9.8 sin 30°) (h / sin 30° - 0)
h = 44.7 m
Using energy:
KE = PE
1/2 mv² = mgh
h = v² / (2g)
h = (29.6)² / (2×9.8)
h = 44.7 m
